For this purpose it is sufficient to show that there exists a set \(B\subseteq {{N}}^{{N}}\) , which violates the perfect subset property.
A set \(B\)  satisfies the perfect subset property if it is either countable or it contains a non-empty perfect subset, which
is a subset that is closed and has no isolated points. We recall that every \(A\subseteq {{N}}^{{N}}\)  into which Cantor space can be continuously embedded contains a perfect subset.
It is a well-known fact that there are so-called Bernstein sets [1]}, which violate the perfect subset property,
at least if we assume the axiom of choice [2]}.
